Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}-x-8y &= 4 \\ x-3y &= 1\end{align*}$
Explanation: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-3y = -x+1$ Divide both sides by $-3$ to isolate $y$ $y = {\dfrac{1}{3}x - \dfrac{1}{3}}$ Substitute this expression for $y$ in the first equation. $-x-8({\dfrac{1}{3}x - \dfrac{1}{3}}) = 4$ $-x - \dfrac{8}{3}x + \dfrac{8}{3} = 4$ Simplify by combining terms, then solve for $x$ $-\dfrac{11}{3}x + \dfrac{8}{3} = 4$ $-\dfrac{11}{3}x = \dfrac{4}{3}$ $x = -\dfrac{4}{11}$ Substitute $-\dfrac{4}{11}$ for $x$ back into the top equation. $+ \dfrac{4}{11}-8y = 4$ $\dfrac{4}{11}-8y = 4$ $-8y = \dfrac{40}{11}$ $y = -\dfrac{5}{11}$ The solution is $\enspace x = -\dfrac{4}{11}, \enspace y = -\dfrac{5}{11}$.